In the circuit the galvanometer `G` shows zero deflection. If the batteries A and B have negligible internal resistance find the value of the resitor `R`

Answer

`100 OMega`

Explanation

Since galvian no meter showns no deffection .hence pot diff across `R = 2` volt .Now the circuit reduces to as in closed loop `ABXCDEFA` we have
`500 l = -2 + 12 = 10 or l = ((10)/(500) = (1)/(50)) A`

in closed loop `ABEFA` we have
`500 l + IR = 12 or 500 xx (1)/(50) + (1)/(50) R = 12`
or `R = 100 Omega`